MCQ 511 Mark
The product of two irrational number is:
- A
- B
- C
- D
sometimes rational and sometimes irrational.
Answer- sometimes rational and sometimes irrational.
Solution:
Consider, the irrational number, $\sqrt{3}.$
Product $=\sqrt{3}\times\sqrt{3}=3,$ which is a rational number.
Consider two irrational numbers, $\sqrt{2}$ and $\sqrt{3}.$
Product $=\sqrt{2}\times\sqrt{3}=\sqrt{6},$ which is an irrational number.
Hence, the product of two irrational numbers are sometimes rational and sometimes irrational.
Hence, the correct opion is (d).
View full question & answer→MCQ 521 Mark
An irrational number between $\frac{1}{7}$ and $\frac{2}{7}$ is :
- A
$\frac{1}{2}\Big(\frac{1}{7}+\frac{2}{7}\Big)$
- B
$\Big(\frac{1}{7}\times\frac{2}{7}\Big)$
- C
$\sqrt{\frac{1}{7}\times\frac{2}{7}}$
- D
Answer- $\sqrt{\frac{1}{7}\times\frac{2}{7}}$
Solution :
An irrational number between a and b is given by $\sqrt{\text{ab}}$
So, an irrational number between $\frac{1}{7}$ and $\frac{2}{7}$ is $\sqrt{\frac{1}{7}\times\frac{2}{7}}$
Hence, the correct answer is option (c).
View full question & answer→MCQ 531 Mark
If $\frac{\sqrt3-1}{\sqrt3+1}=\text{a}-\text{b}\sqrt3,$ then :
Answer- a = 2, b = 1
Solution :
$\frac{\sqrt3-1}{\sqrt3+1}$
Multiplying and dividing by the rationalisation factor of denominator, we get
$\frac{\sqrt3-1}{\sqrt3+1}\times\frac{\sqrt3-1}{\sqrt3-1}$
$=\frac{\big(\sqrt3-1\big)^2}{\big(\sqrt3\big)^2-1^2}$
$=\frac{3-2\sqrt3+1}{3-1}$
$=\frac{4-2\sqrt3}{2}$
$=\frac{2(2-\sqrt3)}{2}$
$=2-\sqrt3$
Comparing with $\text{a}-\text{b}\sqrt3,$ we get a = 2 and b = 1.
Hence, correct option is (a).
View full question & answer→MCQ 541 Mark
Between any two rational numbers there :
- A
- B
is exactly one rational numbers.
- C
are infinitely many rational numbers.
- D
Answer- are infinitely many rational numbers.
Solution :
Options (a), (b) and (d) are incorrect since between two rational numbers there are infinitely many rational and irrational numbers.
Hence, the correct opion is (c).
View full question & answer→MCQ 551 Mark
A number which can neither be expressed as a terminating decimal nor as a repeating decimal is called :
Answer- An irrational number.
Solution :
An irrational number cannot be written in the form of $\frac{\text{p}}{\text{q}}.$
Irrational number can neither be expressed as terminating decimal nor as repeating decimal.
View full question & answer→MCQ 561 Mark
If $\text{x}=\big(7+4\sqrt{3}\big)$ then $\Big(\text{x}+\frac{1}{\text{x}}\Big)=?$
Answer- 14
Solution:
$\text{x}=\big(7+4\sqrt{3}\big)$
$\therefore\frac{1}{\text{x}}=\frac{1}{\big(7+4\sqrt{3}\big)}$
$=\frac{1}{\big(7+4\sqrt{3}\big)}\times\frac{\big(7-4\sqrt{3}\big)}{\big(7-4\sqrt{3}\big)}$
$=\frac{\big(7-4\sqrt{3}\big)}{7^2-\big(4\sqrt{3}\big)^2}$
$=\frac{7-4\sqrt{3}}{49-48}$
$=7-4\sqrt{3}$
$\therefore\Big(\text{x}+\frac{1}{\text{x}}\Big)=\big(7+4\sqrt{3}\big)+\big(7-4\sqrt{3}\big)=14$
Hence, the correct option is (b).
View full question & answer→MCQ 571 Mark
The decimal expansion that a rational number cannot have is:
- A
- B
$0.25\overline{28}$
- C
$0.\overline{2528}$
- D
Answer- 0.5030030003...
Solution:
The decimal expansion of a rational number is either terminating or non-terminating recurring.
The decimal expansion of 0.5030030003... is non-terminating, non-recurring, which is not a property of a rational number.
Hence, the correct opion is (d).
View full question & answer→MCQ 581 Mark
Which of the following numbers is irrational?
Answer- $\sqrt{8}$
Solution:
$\therefore\sqrt{8}=\sqrt{2\times2\times2}=2\sqrt{2}$
View full question & answer→MCQ 591 Mark
$\frac{1}{\sqrt{9}-\sqrt{8}}$ is equal to :
Answer- $3+2\sqrt{2}$
Solution:
After rationalising
$\frac{1}{\sqrt{9}-\sqrt{8}}=\frac{1}{\sqrt{9}-\sqrt{8}}\times\frac{\sqrt{9}+\sqrt{8}}{\sqrt{9}+\sqrt{8}}$
$=\frac{\sqrt{9}+\sqrt{8}}{(\sqrt{9})^2-(\sqrt{8})^2}$
$=\frac{\sqrt{3\times3}+\sqrt{2\times2\times2}}{9-8}$
$=\frac{3+2\sqrt{2}}{1}$
$=3+2\sqrt{2}$
View full question & answer→MCQ 601 Mark
Write the correct answer in the following:
Value of (256)0.16 × (256)0.09 is.
Answer- 4
Solution:
$(256)^{0.16}\times(256)^{0.09}=(256)^{\frac{16}{100}}\times(256)^{\frac{9}{100}}$
$=(256)^{\frac{16}{100}+\frac{9}{100}}\ [\because\text{x}^\text{a}\cdot\text{x}^{\text{b}}=\text{x}^{\text{a+b}}]$
$(256)^{\frac{25}{100}}=(256)^{\frac{1}{4}}$
$=(4^4)^{\frac{1}{4}}=4\ [\because(\text{a}^\text{m})^\text{n}=\text{a}^\text{mn}]$
View full question & answer→MCQ 611 Mark
The value of $\sqrt{\text{p}^{-1}\text{q}}\times\sqrt{\text{q}^{-1}\text{r}}\times\sqrt{\text{r}^{-1}\text{p}}$ is:
Answer- 1
Solution:
$\sqrt{\text{p}^{-1}\text{q}}\times\sqrt{\text{q}^{-1}\text{r}}\times\sqrt{\text{r}^{-1}\text{p}}$
$=\sqrt{\frac{\text{p}}{\text{q}}}\times\sqrt{\frac{\text{r}}{\text{q}}}\times\sqrt{\frac{\text{p}}{\text{r}}}$
$=\sqrt{\frac{\text{q}}{\text{p}}\times\frac{\text{r}}{\text{q}}\times\frac{\text{p}}{\text{r}}}$
$=1$
View full question & answer→MCQ 621 Mark
The simplest for of $0.\overline{32}$ is:
- A
$\frac{16}{45}$
- B
$\frac{32}{99}$
- C
$\frac{29}{90}$
- D
Answer- $\frac{29}{90}$
Solutions:
Let $\text{x}=0.\overline{32}$
Then, $\text{x}=0.3222 \ ...(\text{i})$
$\therefore10\text{x}=3.222 \ ...(\text{ii})$
and $100\text{x}=32.222 \ ...(\text{iii})$
On subtracting (ii) from (iii), we get
$90\text{x}=29$
$\Rightarrow\text{x}=\frac{29}{90}$
Hence, the correct option is (c).
View full question & answer→MCQ 631 Mark
The value of (xa-b)a+b × (xb-c)b+c × (xc-a)c+a is:
Answer- 1
Solution:
(xa-b)a+b × (xb-c)b+c × (xc-a)c+a
$\Rightarrow\text{x}^{\text{a}^{2}-\text{b}^2}\times\text{x}^{\text{b}^2-\text{c}^2}\times\text{x}^{\text{c}^2-\text{a}^2}$
$\Rightarrow\text{x}^{\text{a}^2-\text{b}^2+\text{b}^2-\text{c}^2+\text{c}^2-\text{a}^2}$
$\Rightarrow\text{x}^0=1$
View full question & answer→MCQ 641 Mark
$\sqrt[5]{6}\times\sqrt[5]{6}$ is equal to:
- A
$\sqrt[5]{36}$
- B
$\sqrt[5]{6\times0}$
- C
$\sqrt[5]{6}$
- D
$\sqrt[5]{12}$
Answer- $\sqrt[5]{36}$
Solution:
$\sqrt[5]{6}=(6)^{\frac{1}{5}}$
So $\sqrt[5]{6}\times\sqrt[5]{6}=(6)^{\frac{1}{5}}\times(6)^{\frac{1}{5}}$
$=(6\times6)^{\frac{1}{5}}$
$(36)^{\frac{1}{5}}$
$=\sqrt[5]{36}$
Hence, correct option is (a).
View full question & answer→MCQ 651 Mark
$\frac{1}{\sqrt9-\sqrt8}$ is equal to:
- A
$3+2\sqrt2$
- B
$\frac{1}{3+2\sqrt2}$
- C
$3-2\sqrt2$
- D
$\frac{3}{2}-\sqrt2$
Answer- $3+2\sqrt2$
Solution:
$\frac{1}{\sqrt9-\sqrt8}$
$=\frac{1}{\sqrt9-\sqrt8}\times\frac{{\sqrt9+\sqrt8}}{{\sqrt9+\sqrt8}}$
$\frac{{\sqrt9+\sqrt8}}{\big(\sqrt9\big)^2-\big(\sqrt8\big)^2}$
$=\frac{{\sqrt9+\sqrt8}}{9-8}$
$={\sqrt9+\sqrt8}$
$=3+2\sqrt2$
Hence, correct option is (a).
View full question & answer→MCQ 661 Mark
Write the correct answer in the following:
Which of the following is equal to x?
- A
$\text{x}^{\frac{12}{7}}-\text{x}^{\frac{5}{7}}$
- B
$\sqrt[12]{(\text{x}^4)^{\frac{1}{3}}}$
- C
$(\sqrt{\text{x}^3})^{\frac{2}{3}}$
- D
$\text{x}^{\frac{12}{7}}\times\text{x}^{\frac{7}{12}}$
Answer- $(\sqrt{\text{x}^3})^{\frac{2}{3}}$
Solution:
- $\text{x}^{\frac{12}{7}}-\text{x}^{\frac{5}{7}}\neq\text{x}$
- $\sqrt[12]{\text{(x}^4})^{\frac{1}{3}}=\sqrt[12]{\text{x}^{4\times\frac{1}{3}}}=\Big(\text{x}^\frac{4}{3}\Big)^\frac{1}{12}=\text{x}^{\frac{4}{3}\times\frac{1}{12}}=\text{x}^\frac{1}{9}\neq\text{x}$
- $\Big((\text{x}^3)^\frac{1}{2}\Big)^\frac{2}{3}=\text{(x)}^{\frac{3}{2}\times\frac{2}{3}}=\text{x}^1=\text{x}$
- $\text{x}^{\frac{12}{7}}\times\text{x}^\frac{7}{12}=\text{x}^{\frac{12}{7}+\frac{7}{12}}=\text{x}^\frac{193}{84}\neq\text{x}$
Hence, (c) is the correct answer.
View full question & answer→MCQ 671 Mark
If $\text{x}^{-2}=64,$ then $\text{x}^{\frac{1}{3}}+\text{x}^0=$
- A
$2$
- B
$3$
- C
$\frac{3}{2}$
- D
$\frac{2}{3}$
Answer- $\frac{3}{2}$
Solution:
We have to find the value of $\text{x}^{\frac{1}{3}}+\text{x}^0$ if $\text{x}^{-2}=64$
Consider,
$\text{x}^{-2}=2^6$
$\frac{1}{\text{x}^2}=2^6$
Multiply $\frac{1}{2}$ on both sides of powers we get
$\frac{1}{\text{x}^{2\times\frac{1}{2}}}=2^{6\times\frac{1}{6}}$
$\frac{1}{\text{x}}=2^3$
$\frac{1}{\text{x}}=\frac{8}{1}$
By taking reciprocal on both sides we get,
$\frac{1}{8}=\text{x}$
Substituting $\frac{1}{8}$ in $\text{x}^{\frac{1}{3}}+\text{x}^0$ we get
$=\Big(\frac{1}{8}\Big)^{\frac{1}{3}}+\Big(\frac{1}{8}\Big)^0$
$=\Big(\frac{1}{2^2}\Big)^{\frac{1}{3}}+\Big(\frac{1}{8}\Big)^0$
$=\frac{1}{2^{3\times\frac{1}{3}}}+1$
$=\frac{1}{2^1}+1$
$=\frac{1}{2}+1$
By taking least common multiply we get
$=\frac{1}{2}+\frac{1\times2}{1\times2}$
$=\frac{1}{2}+\frac{2}{2}$
$=\frac{1+2}{2}$
$=\frac{3}{2}$
Hence the correct choice is c.
View full question & answer→MCQ 681 Mark
Which one of the following statements is true?
- A
The sum of two irrational numbers is always an irrational number.
- B
The sum of two irrational numbers is always a rational number.
- C
The sum of two irrational numbers may be a rational number or an irrational number.
- D
The sum of two irrational numbers is always an integer.
Answer- The sum of two irrational numbers may be a rational number or an irrational number.
Solution:
If two irrational numbers i.e. $\sqrt{2},\sqrt{5},2+\sqrt{3},2-\sqrt{3}$ etc. are added it is not necessary that sum comes out to be an irrational number always, or a rational nnumber always, or a rational number always...
Since $\sqrt{2}+\sqrt{5}=$ an irrational number
$2+\sqrt{\not\text{3}}+2-\sqrt{\not\text{3}}=4=$ a rational number
So we see that $\sqrt{2}$ and $\sqrt{5}$ are irrational numbers, and their sum is also irrational.
But $2+\sqrt{3}$ and $2-\sqrt{3}$ are also irrational numbers, and their sum is rational number '4'.
So sum of two irrational numbers can be either an irrational number or a rational number depending which numbers are being added.
So options (a) and (b) are totally wrong, because they are not 'always' true.
Option (c) is correcrt because sum can be either irrational or rational and option (c) is verifying this statement.
Option (d) - again it is not always true, if we add two irrational numbers like $2+\sqrt{3}$ and $2-\sqrt{3}.$
Sum is an integer = 4, but if we add $\sqrt{3}$ and $\sqrt{3},$ sum is $2\sqrt{3}$ which is not an integer but again an irrational number.
So option (d) is also incorrect.
Hence, correct option is (c).
View full question & answer→MCQ 691 Mark
The simplest for of $1.\overline{6}$ is:
- A
$\frac{833}{500}$
- B
$\frac{8}{5}$
- C
$\frac{5}{3}$
- D
Answer- $\frac{5}{3}$
Solution:
Let $\text{x}=1.\overline{6}$
$\Rightarrow\text{x}=1.666 \ ...(\text{i})$
$\therefore10\text{x}=16.666 \ ...(\text{ii})$
On subtracting (i) from (ii), we get
$9\text{x}=15$
$\Rightarrow\text{x}=\frac{15}{9}=\frac{5}{3}$
Hence, the correct option is (c).
View full question & answer→MCQ 701 Mark
If $\sqrt{2}=1.41$ then $\frac{1}{\sqrt{2}}=?$
Answer- 0.705
Solution:
$\frac{1}{\sqrt{2}}=\frac{1}{\sqrt{2}}\times\frac{\sqrt{2}}{\sqrt{2}}$
$=\frac{\sqrt{2}}{2}$
$=\frac{1.41}{2}$
$=0.705$
Hence, the correct option is (c).
View full question & answer→MCQ 711 Mark
$(1296)\frac{-1}{4}=$
- A
$-\frac{1}{6}$
- B
- C
- D
$\frac{1}{6}$
Answer- $\frac{1}{6}$
Solution:
$(1296)\frac{-1}{4}=(6^4)^{\frac{-1}{4}}$
$=(6)^{-1}=\frac{1}{6}$
View full question & answer→MCQ 721 Mark
Directions: In the following questions, the Assertions (A) and Reason(s) (R) have been put forward. Read both the statements carefully and choose the correct alternative from the following:
Assertion: Sum of two irrational numbers $2+\sqrt3$ is an irrational number.
Reason: Sum of a rational number and an irrational numbers is always an irrational number.
- A
Both Assertion and Reason are correct and Reason is the correct explanation for Assertion.
- B
Both Assertion and Reason are correct and Reason is not the correct explanation for Assertion.
- C
Assertion is true but the reason is false.
- D
Both assertion and reason are false.
Answer - Both Assertion and Reason are correct and Reason is the correct explanation for Assertion.
View full question & answer→MCQ 731 Mark
$\frac{1}{(3+2\sqrt{2})}=\text{?}$
Answer- $(3-2\sqrt{2})$
Solution:
$\frac{1}{(3+2\sqrt{2})}$
$=\frac{3-2\sqrt{2}}{(3+2\sqrt{2})(3-2\sqrt{2})}$
$=(3-2\sqrt{2})$
View full question & answer→MCQ 741 Mark
If 2x = 4y and $\frac{1}{2\text{x}}+\frac{1}{4\text{y}}+\tfrac{1}{4\text{z}}=4$ then the value of x is:
- A
$\frac{3}{4}$
- B
$\frac{4}{3}$
- C
$\frac{16}{7}$
- D
$\frac{7}{16}$
Answer- $\frac{7}{16}$
Solution:
2x = 4y = 8z
⇒ 2x = 22y = 23z
⇒ x = 2y = 3z
$\Rightarrow\text{y}=\frac{\text{x}}{2}$ and $\Rightarrow\text{z}=\frac{\text{x}}{3}$
So, $\frac{1}{2\text{x}}+\tfrac{1}{4\text{y}}+\frac{1}{4\text{z}}=4$
$\Rightarrow\frac{1}{2\text{x}}+\frac{2}{4\text{x}}+\frac{3}{4\text{x}}=4$
$\Rightarrow\frac{7}{4\text{x}}=4$
$\Rightarrow\text{x}=\frac{7}{16}$
View full question & answer→MCQ 751 Mark
The product of the square root of x with the cube root of x is:
- A
Cube root of the square root of x.
- B
Sixth root of the fifth power of x.
- C
Fifth root of the sixth power of x.
- D
Answer- Sixth root of the fifth power of x.
Solution:
We have to find the product (say L) of the square root of x with the cube root of x is.
So,
$\text{L}=\sqrt[2]{\text{x}}\times\sqrt[3]{\text{x}}$
$=\text{x}^\frac{1}{2}\times\text{x}^\frac{1}{3}$
$=\text{x}^{\frac{1}{2}+\frac{1}{3}}$
$=\text{x}^{\frac{1\times3}{2\times3}+\frac{1\times2}{3\times2}}$
$=\text{x}^{\frac{3+2}{6}}=\text{x}^\frac{5}{6}$
The product of the square root of x with the cube root of x is $\text{x}^{\frac{5}{6}}$
Hence the correct alternative is b.
View full question & answer→MCQ 761 Mark
$(\frac{125}{216})^{\frac{-1}{3}}=$
- A
$\frac{5}{6}$
- B
$\frac{6}{5}$
- C
- D
Answer- $\frac{6}{5}$
Solution:
$(\frac{125}{216})^{\frac{-1}{3}}$
$\Rightarrow(\frac{5}{6})^3\times\frac{-1}{3}$
$\Rightarrow(\frac{5}{6})^{-1}$
$\Rightarrow\frac{6}{5}$
View full question & answer→MCQ 771 Mark
On simplification, the expression $\frac{5^{\text{n}+2}-6\times5^{\text{n}+1}}{13\times5^{\text{n}}-2\times5^{\text{n}+1}}$ equals :
- A
$\frac{5}{3}$
- B
$-\frac{5}{3}$
- C
$\frac{3}{5}$
- D
$-\frac{3}{5}$
Answer- $-\frac{5}{3}$
Solution :
$\frac{5^{\text{n}+2}-6\times5^{\text{n}+1}}{13\times5^{\text{n}}-2\times5^{\text{n}+1}}$
$=\frac{5^{\text{n}+1}(5-6)}{5^{\text{n}}(13-2\times5)}$
$=\frac{5^{\text{n}}\times5\times(-1)}{5^{\text{n}}(13-10)}$
$=-\frac{5}{3}$
Hence, the correct option is (b).
View full question & answer→MCQ 781 Mark
The value of $\frac{4\sqrt{12}}{12\sqrt{27}}$ is:
- A
$\frac{4}{9}$
- B
$\frac{8}{9}$
- C
$\frac{2}{9}$
- D
$\frac{1}{9}$
Answer- $\frac{2}{9}$
Solution:
$\frac{4\sqrt{12}}{12\sqrt{27}}$
$=\frac{4\sqrt{4\times3}}{12\sqrt{9\times3}}$
$=\frac{8\sqrt{3}}{36\sqrt{3}}$
$=\sqrt{\frac{(\sqrt{2}-1)^2}{2-1}}$
$=\sqrt{(\sqrt{2}-1)^2}$
$=1.414-1$
$=0.414$
$=\frac{2}{9}$
View full question & answer→MCQ 791 Mark
Write the correct answer in the following:
The number obtained on rationalising the denominator of $\frac{1}{\sqrt{7}-2}$ is
- A
$\frac{\sqrt{7}+2}{3}$
- B
$\frac{\sqrt{7}-2}{3}$
- C
$\frac{\sqrt{7}+2}{5}$
- D
$\frac{\sqrt{7}+2}{45}$
Answer- $\frac{\sqrt{7}+2}{3}$
Solution:
$\frac{1}{\sqrt{7}-2}=\frac{1}{\sqrt{7}-2}\times\frac{\sqrt{7}+2}{\sqrt{7}+2}$
$\frac{\sqrt{7}+2}{7-4}=\frac{\sqrt{7}+2}{3}$
Hence, (a) is the correct answer.
View full question & answer→MCQ 801 Mark
If $\text{x}^{\frac{1}{12}}=49^{\frac{1}{24}},$ then the value of x is:
Answer- 7
Solution:
$\text{x}^{\frac{1}{12}}=49^{\frac{1}{24}},$
$\Rightarrow\text{x}^{\frac{1}{12}}=7\frac{2}{24}=7^\frac{1}{12}$
Equating both, x = 7
View full question & answer→MCQ 811 Mark
If $0<\text{y}<\text{x},$ which statement must be true?
- A
$\sqrt{\text{x}}-\sqrt{\text{y}}=\sqrt{\text{x}-\text{y}}$
- B
$$$\sqrt{\text{x}}+\sqrt{\text{x}}=\sqrt{2\text{x}}$
- C
$\text{x}\sqrt{\text{y}}=\text{y}\sqrt{\text{x}}$
- D
$\sqrt{\text{xy}}=\sqrt{\text{x}}\sqrt{\text{y}}$
Answer- $\sqrt{\text{xy}}=\sqrt{\text{x}}\sqrt{\text{y}}$
Solution:
We have to find which statement must be true?
Given $0<\text{y}<\text{x},$
Option (a):
Left hand side:
$\sqrt{\text{x}}-\sqrt{\text{y}}=\sqrt{\text{x}\text{}}=\sqrt{\text{y}}$
Right Hand side:
$\sqrt{\text{x}-\text{y}}=\sqrt{\text{x}-\text{y}}$
Left hand side is not equal to right hand side
The statement is wrong.
Option (b):
$\sqrt{\text{x}}+\sqrt{\text{x}}=\sqrt{2\text{x}}$
Left hand side:
$\sqrt{\text{x}}+\sqrt{\text{x}}=2\sqrt{\text{x}}$
Right Hand side:
$\sqrt{2\text{x}}=\sqrt{2\text{x}}$
Left hand side is not equal to right hand side
The statement is wrong.
Option (c):
$\text{x}\sqrt{\text{y}}=\text{y}\sqrt{\text{x}}$
Left hand side:
$\text{x}\sqrt{\text{y}}=\text{x}\sqrt{\text{y}}$
Right Hand side:
$\text{y}\sqrt{\text{x}}=\text{y}\sqrt{\text{x}}$
Left hand side is not equal to right hand side
The statement is wrong.
Option (d):
$\sqrt{\text{xy}}=\sqrt{\text{x}}\sqrt{\text{y}}$
Left hand side:
$\sqrt{\text{xy}}=\sqrt{\text{xy}}$
Right Hand side:
$\sqrt{\text{x}}\sqrt{\text{y}}=\sqrt{\text{x}}\times\sqrt{\text{y}}$
$=\sqrt{\text{xy}}$
Left hand side is equal to right hand side
The statement is true.
Hence the correct choice is d.
View full question & answer→MCQ 821 Mark
Write the correct answer in the following:
$\sqrt{10}\times\sqrt{15}$ is equal to.
- A
$6\sqrt{5}$
- B
$5\sqrt{6}$
- C
$\sqrt{25}$
- D
$10\sqrt{5}$
Answer- $5\sqrt{6}$
Solution :
$\sqrt{10}, \sqrt{15}=\sqrt{2.5}\sqrt{3.5}=\sqrt{2}\sqrt{5}\sqrt{3}\sqrt{5}=5\sqrt{6}$
View full question & answer→MCQ 831 Mark
If 8x+1 = 64, what is the value of 32x+1?
Answer- 27
Solution :
We have to find the value of 32x+1 provided 8x+1 = 64
So,
23(x+1) = 64
23x+3 = 26
Equating the exponents we get
3x + 3 = 6
3x = 6 - 3
3x = 3
$\text{x}=\frac{3}{3}$
x = 1
By substitute in 32x+1 we get
= 32×1+1
= 32+1
= 33
= 27
The real value of 32x+1 is 27
Hence the correct choice is d.
View full question & answer→MCQ 841 Mark
If $\text{x}=2+\sqrt{3}$ then $\Big(\text{x}+\frac{1}{\text{x}}\Big)$ equals :
- A
$-2{\sqrt{3}}$
- B
- C
- D
$4-2\sqrt{3}$
Answer- 4
Solution :
$\text{x}=2+\sqrt{3}$
$\therefore\frac{1}{\text{x}}=\frac{1}{2+\sqrt{3}}$
$=\frac{1}{2+\sqrt{3}}\times\frac{2-\sqrt{3}}{2-\sqrt{3}}$
$=\frac{2-\sqrt{3}}{2^2-\big(\sqrt{3}\big)^2}$
$=\frac{2-\sqrt{3}}{4-3}$
$=2-\sqrt{3}$
$\therefore\Big(\text{x}+\frac{1}{\text{x}}\Big)=\big(2+\sqrt{3}\big)+\big(2-\sqrt{3}\big)=4$
Hence, the correct option is (c).
View full question & answer→MCQ 851 Mark
The difference of two irrational numbers is.
- A
- B
- C
- D
Either irrational or rational.
Answer- Either irrational or rational.
Solution :
Difference of two irrationals need not be an irrational.
Example:- each one of $(5\div\sqrt{2})$ and $(5-\sqrt{2})$ is irrational,
But, $(5+\sqrt{2})-(3+\sqrt{2})=2,$ which is rational.
View full question & answer→MCQ 861 Mark
$\sqrt{10}\times\sqrt{15}$ is equal to :
- A
$5\sqrt6$
- B
$6\sqrt5$
- C
$\sqrt{30}$
- D
$\sqrt{25}$
Answer- $5\sqrt6$
Solution :
10 = 5 × 2
15 = 5 × 3
$\therefore\sqrt{10}\times\sqrt{15}=\sqrt{5\times2}\times\sqrt{5\times3}$
$=\sqrt{5}\times\sqrt{2}\times\sqrt{5}\times\sqrt3$
$=\big(\sqrt5\times\sqrt5\big)\times\sqrt2\times\sqrt3$
$=5\sqrt6$
Hence, correct option is (a).
View full question & answer→MCQ 871 Mark
If $\text{x}=3+\sqrt{8},$ then the value of $(\text{x}^2+\frac{1}{\text{x}^2})$ is:
Answer- 34
Solution:
Given, $(\text{3}+\sqrt{8})$
$\frac{1}{\text{x}}=\frac{1}{(3+\sqrt{8})}=\frac{1}{(3+\sqrt{8})}\times\frac{(3-\sqrt{8})}{(3-\sqrt{8})}$
$=\frac{(3-\sqrt{8})}{(3^2-(\sqrt{8})^2)}=\frac{3+\sqrt{8}}{9-8}=(3-\sqrt{8})$
$(\text{x}+\frac{1}{\text{x}})=(3+\sqrt{8})+(3-\sqrt{8})=6$
$\Rightarrow(\text{x}+\frac{1}{\text{x}})^2=6^2=36$
$\Rightarrow(\text{x}^2+\frac{1}{\text{x}^2})+2\times\text{x}\times\frac{1}{\text{x}}=36$
$\Rightarrow(\text{x}^2+\frac{1}{\text{x}^2})+2=36$
$\Rightarrow(\text{x}+\frac{1}{\text{x}^2})=36-2=34$
View full question & answer→MCQ 881 Mark
If $\sqrt{2}=1.414$ then $\sqrt{\frac{\big(\sqrt{2}-1\big)}{\big(\sqrt{2}+1\big)}}=?$
Answer- 0.414
Solution:
$\sqrt{\frac{\big(\sqrt{2}-1\big)}{\big(\sqrt{2}+1\big)}}=\sqrt{\frac{\big(\sqrt{2}-1\big)}{\big(\sqrt{2}+1\big)}\times\frac{\big(\sqrt{2}-1\big)}{\big(\sqrt{2}-1\big)}}$
$=\sqrt{\frac{\big(\sqrt{2}-1\big)^2}{\big(\sqrt{2}\big)^2-(1)^2}}$
$=\sqrt{\frac{\big(\sqrt{2}-1\big)^2}{1}}$
$=\sqrt{\big(\sqrt{2}-1\big)^2}$
$=\sqrt{2}-1$
$=1.414-1$
$=0.414$
Hence, the correct option is (c).
View full question & answer→MCQ 891 Mark
If 'm' is a positive integer which is not a perfect square, then $\sqrt{\text{m}}$ is:
Answer- An irrational number.
Solution:
In $\sqrt{\text{m}}$ ,if m is not perfect square ,then the value of $\sqrt{\text{m}}$ will be $\text{m}^\frac{1}{2}.$
But if it is perfect square then value of $\sqrt{\text{m}}=$ some integer,
Example $\sqrt{4}=2$
View full question & answer→MCQ 901 Mark
Answer- Always a whole number.
Solution:
All natural number together with zero form the collection of all whole number.
So,
- Every natural number is a whole number.
- Zero is a whole number which is not a natural number.
View full question & answer→MCQ 911 Mark
Value of $\sqrt[4]{(81)^{-2}}$ is:
- A
$\frac{1}{81}$
- B
$\frac{1}{3}$
- C
$\frac{1}{9}$
- D
$9$
Answer- $\frac{1}{9}$
Solution:
$\sqrt[4]{(81)^{-2}}$
$=\sqrt[4]{\frac{1}{(81)^2}}$
$=\sqrt[4]{\frac{1}{(9^2)^2}}$
$=\sqrt[4]{\frac{1}{9^4}}$
$=\Big(\frac{1}{9}\Big)^{4\times\frac{1}{4}}$
$=\frac{1}{9}$
View full question & answer→MCQ 921 Mark
$16\sqrt{134}\div9\sqrt{52}$ is equal to :
- A
$\frac{3}{9}$
- B
$\frac{9}{8}$
- C
$\frac{8}{9}$
- D
Answer- $\frac{8}{9}$
Solution :
$16\sqrt{134}\div9\sqrt{52}$
$\frac{16\sqrt{13}}{9\sqrt{52}}=\frac{16}{9}\times\sqrt{\frac{13}{52}}=\frac{16}{9}\times\frac{1}{2}$
$=\frac{8}{9}$
View full question & answer→MCQ 931 Mark
Which of the following is true statement ?
- A
Every real number is either rational or irrational.
- B
The product of two irrational numbers is an irrational number.
- C
The sum of two irrational numbers is an irrational number.
- D
Every real number is always rational.
Answer- Every real number is either rational or irrational.
Solution :
Consider, $(2+\sqrt{3})$ and $(2-\sqrt{3})$ which are two irrational number.
$(2+\sqrt{3})+(2-\sqrt{3})=4,$ which is a rational number.
Consider, $\sqrt{3}$ and $\frac{1}{\sqrt{3}}$ which is a rational number.
Every real number can either be a rational number or an irrational number.
View full question & answer→MCQ 941 Mark
If $\text{x}=3+2\sqrt{2},$ than the value of $\text{x}+\frac{1}{\text{x}}$ is:
Answer- 6
Solution:
$\text{x}+\frac{1}{\text{x}}$
$\Rightarrow\frac{\text{x}^2+1}{\text{x}}$
Put the value of x,
$\Rightarrow\frac{(3+2\sqrt{2})^2+1}{3+2\sqrt{2}}$
$\Rightarrow\frac{9+8+12\sqrt{2}+1}{3+2\sqrt{2}}$
$\Rightarrow\frac{18+12\sqrt{2}}{3+2\sqrt{2}}$
$\Rightarrow\frac{6(3+2\sqrt{2})}{3+2\sqrt{2}}$
$\Rightarrow6$
View full question & answer→MCQ 951 Mark
Which of the following statement is true?
- A
$\sqrt{\text{a}+\text{b}}=\sqrt{\text{a}}+\sqrt{\text{b}}$
- B
$\sqrt{\text{a}-\text{b}}=\sqrt{\text{a}}-\sqrt{\text{b}}$
- C
$\sqrt[3]{\text{ab}}=\sqrt{\text{a}}\times\sqrt{\text{b}}$
- D
$\sqrt{\text{ab}}=\sqrt{\text{a}}\times\sqrt{\text{b}}$
Answer- $\sqrt{\text{ab}}=\sqrt{\text{a}}\times\sqrt{\text{b}}$
Solution:
When we multiply two different root number then only number is multiplied not the root because we have both number power equal,
$\sqrt{\text{a}}\times\sqrt{\text{b}}=\sqrt{\text{ab}}$
or, $\text{a}\frac{1}{2}\times\text{b}\frac{1}{2}=(\text{ab})\frac{1}{2}$
View full question & answer→MCQ 961 Mark
An irrational number between s and 6 is:
- A
$\sqrt{5\times6}$
- B
$\sqrt{5+6}$
- C
$\sqrt{5-6}$
- D
Answer- $\sqrt{5\times6}$
Solution:
We know that, If a and b are two distinct positive rational numbers such that ab is not a perfect square of a rational number, Then $\sqrt{\text{ab}}$ is an irrational number lying between a and b,
Here also we have 5 and 6 two distinct rational numbers and 5 × 6 = 30 is not a perfect square,
So irrational number betweens and 6 $=\sqrt{5\times6}$
View full question & answer→MCQ 971 Mark
The decimal form of $\frac{\text{1}}{999}$ is :
- A
$0.0\overline{01}$
- B
$0.\overline{001}$
- C
$0.999$
- D
$0.00\overline{1}$
Answer- $0.\overline{001}$
Solution:
When we divide 1 by 999 result is 0.001001001001001.....
So, $0.\overline{001}=\frac{1}{999}$
View full question & answer→MCQ 981 Mark
The value of $15\sqrt{15}\div3\sqrt{5}$ is:
- A
- B
- C
$5\sqrt{3}$
- D
$3\sqrt{5}$
Answer- $5\sqrt{3}$
Solution:
$\frac{15\sqrt{15}}{3\sqrt{5}}$
$=\frac{(3\times5)\sqrt{3}\times\sqrt{5}}{3\sqrt{3}}$
$=5\sqrt{3}$
View full question & answer→MCQ 991 Mark
$\sqrt{8}+2\sqrt{32}-5\sqrt{2}$ is equal to:
- A
- B
$\sqrt{32}$
- C
$\sqrt{8}$
- D
$5\sqrt{2}$
Answer- $5\sqrt{2}$
Solution:
$\sqrt{8}+2\sqrt{32}-5\sqrt{2}$
$\Rightarrow2\sqrt{2}+2\times4\sqrt{2}-5\sqrt{2}$
$\Rightarrow10\sqrt{2}-5\sqrt{2}$
$\Rightarrow5\sqrt{2}$
View full question & answer→MCQ 1001 Mark
$2^{\frac{4}{3}}$ is same as.
- A
$\sqrt[3]{2^4}$
- B
- C
$\sqrt[3]{4}$
- D
$\sqrt[4]{2^3}$
Answer- $\sqrt[3]{2^4}$
Solution:
$\sqrt[3]{2^4}$
$=(2^4)^{\frac{1}{3}}$
$=2^{\frac{4}{3}}$
View full question & answer→